A simple harmonic motion is produced when a restoring force proportional to the displacement acts on a particle. If the particle is acted upon by two separate forces each of which can produce a simple harmonic motion, the resultant motion of the particle is a combination of two simple harmonic motions.

Let \(\overrightarrow{r_1}\) denote the position of the particle at time \(t\) if the force \(\vec{F}_1\) alone acts on it. Similarly, let \(\overrightarrow{r_2}\) denote the position at time \(t\) if the force \(\vec{F}_2\) alone acts on it. Newton’s second law gives,
\(
\begin{aligned}
& m \frac{d^2 \overrightarrow{r_1}}{d t^2}=\vec{F}_1
\end{aligned}
\)
and
\(
m \frac{d^2 \overrightarrow{r_2}}{d t^2}=\overrightarrow{F_2}
\)
Adding them,
\(
m \frac{d^2 \vec{r}_1}{d t^2}+m \frac{d^2 \vec{r}_2}{d t^2}=\vec{F}_1+\vec{F}_2
\)
\(
\text { or, } \quad m \frac{d^2}{d t^2}\left(\overrightarrow{r_1}+\overrightarrow{r_2}\right)=\vec{F}_1+\vec{F}_2 \dots(i)
\)
But \(\vec{F}_1+\vec{F}_2\) is the resultant force acting on the particle and so the position \(\vec{r}\) of the particle when both the forces act, is given by
\(
m \frac{d^2 \vec{r}}{d t^2}=\vec{F}_1+\vec{F}_2 \dots(ii)
\)
Comparing (i) and (ii) we can show that
\(
\begin{aligned}
& \vec{r}=\overrightarrow{r_1}+\overrightarrow{r_2} \\
& \vec{u}=\vec{u}_1+\vec{u}_2
\end{aligned}
\)
\(
\text { and } \quad \vec{u}=\vec{u}_1+\vec{u}_2
\)
if these conditions are met at \(t=0\).
Thus, if two forces \(\vec{F}_1\) and \(\vec{F}_2\) act together on a particle, its position at any instant can be obtained as follows. Assume that only the force \(\vec{F}_1\) acts and find the position \(\overrightarrow{r_1}\) at that instant. Then assume that only the force \(\vec{F}_2\) acts and find the position \(\vec{r}_2\) at that same \(\overrightarrow{r_1}\) and \(\overrightarrow{r_2}\).

Composition of two Simple Harmonic Motions in Same Direction

Suppose two forces act on a particle, the first alone would produce a simple harmonic motion given by
\(
x_1=A_1 \sin \omega t
\)
and the second alone would produce a simple harmonic motion given by
\(
x_2=A_2 \sin (\omega t+\delta)
\)
Both the motions are along the \(x\)-direction. The amplitudes may be different and their phases differ by \(\delta\). Their frequency is assumed to be same. The resultant position of the particle is then given by
\(
\begin{aligned}
x & =x_1+x_2 \\
& =A_1 \sin \omega t+A_2 \sin (\omega t+\delta) \\
& =A_1 \sin \omega t+A_2 \sin \omega t \cos \delta+A_2 \cos \omega t \sin \delta \\
& =\left(A_1+A_2 \cos \delta\right) \sin \omega t+\left(A_2 \sin \delta\right) \cos \omega t
\end{aligned}
\)
\(
\begin{aligned}
& =C \sin \omega t+D \cos \omega t \\
& =\sqrt{C^2+D^2}\left[\frac{C}{\sqrt{C^2+D^2}} \sin \omega t+\right. \\
& \left.\qquad \frac{D}{\sqrt{C^2+D^2}} \cos \omega t\right] \ldots(i)
\end{aligned}
\)
where \(C=A_1+A_2 \cos \delta\) and \(D=A_2 \sin \delta\).
Now \(\frac{C}{\sqrt{C^2+D^2}}\) and \(\frac{D}{\sqrt{C^2+D^2}}\) both have magnitudes
less than 1 and the sum of their squares is 1 . Thus, we can find an angle \(\varepsilon\) between 0 and \(2 \pi\) such that
\(
\sin \varepsilon=\frac{D}{\sqrt{C^2+D^2}} \text { and } \cos \varepsilon=\frac{C}{\sqrt{C^2+D^2}} .
\)
Equation (i) then becomes
\(
x=\sqrt{C^2+D^2}(\cos \varepsilon \sin \omega t+\sin \varepsilon \cos \omega t)
\)
or, \(\quad x=A \sin (\omega t+\varepsilon) \dots(12.26)\)
where
\(
\begin{aligned}
A & =\sqrt{C^2+D^2} \\
& =\sqrt{\left(A_1+A_2 \cos \delta\right)^2+\left(A_2 \sin \delta\right)^2} \\
& =\sqrt{A_1^2+2 A_1 A_2 \cos \delta+A_2^2 \cos ^2 \delta+A_2^2 \sin ^2 \delta} \\
& =\sqrt{A_1^2+2 A_1 A_2 \cos \delta+A_2^2} \dots(12.27)
\end{aligned}
\)
and
\(
\tan \varepsilon=\frac{D}{C}=\frac{A_2 \sin \delta}{A_1+A_2 \cos \delta} \dots(12.28)
\)
Equation (12.26) shows that the resultant of two simple harmonic motions along the same direction is itself a simple harmonic motion. The amplitude and phase of the resultant simple harmonic motion depend on the amplitudes of the two-component simple harmonic motions as well as the phase difference between them.

Amplitude of The Resultant Simple Harmonic Motion

The amplitude of the resultant simple harmonic motion is given by equation (12.27),
\(
A=\sqrt{A_1^2+2 A_1 A_2 \cos \delta+A_2^2} .
\)
If \(\delta=0\), the two simple harmonic motions are in phase
\(
A=\sqrt{A_1^2+2 A_1 A_2+A_2^2}=A_1+A_2 .
\)
The amplitude of the resultant motion is equal to the sum of the amplitudes of the individual motions. This is the maximum possible amplitude.

If \(\delta=\pi\), the two simple harmonic motions are out of phase and
\(
A=\sqrt{A_1^2-2 A_1 A_2+A_2^2}=A_1-A_2 \text { or } A_2-A_1 \text {. }
\)
As the amplitude is always positive we can write \(A=\left|A_1-A_2\right|\). If \(A_1=A_2\) the resultant amplitude is zero and the particle does not oscillate at all.

For any value of \(\delta\) other than 0 and \(\pi\) the resultant amplitude is between \(\left|A_1-A_2\right|\) and \(A_1+A_2\).

Example 1: Find the amplitude of the simple harmonic motion obtained by combining the motions
\(x_1=\left(2.0 \mathrm{~cm}\right) \sin \omega t\)
and
\(x_2=\left(2.0 \mathrm{~cm}\right) \sin (\omega t+\pi / 3)\).

Solution:

The two equations given represent simple harmonic motions along \(X\)-axis with amplitudes \(A_1=2 \cdot 0 \mathrm{~cm}\) and \(A_2=2 \cdot 0 \mathrm{~cm}\). The phase difference between the two simple harmonic motions is \(\pi / 3\). The resultant simple harmonic motion will have an amplitude \(A\) given by
\(
\begin{aligned}
A & =\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \delta} \\
& =\sqrt{(2 \cdot 0 \mathrm{~cm})^2+(2 \cdot 0 \mathrm{~cm})^2+2(2 \cdot 0 \mathrm{~cm})^2 \cos \frac{\pi}{3}} \\
& =3.5 \mathrm{~cm} .
\end{aligned}
\)

Vector Method of Combining Two Simple Harmonic Motions

There is a very useful method to remember the equations of resultant simple harmonic motion when two simple harmonic motions of the same frequency and in the same direction combine. Suppose the two individual motions are represented by

\(
\begin{array}{ll}
& x_1=A_1 \sin \omega t \\
\text { and } & x_2=A_2 \sin (\omega t+\delta) .
\end{array}
\)

Let us for a moment represent the first simple harmonic motion by a vector of magnitude \(A_1\) and the second simple harmonic motion by another vector of magnitude \(A_2\). We draw these vectors in figure (12.16). The vector \(A_2\) is drawn at an angle \(\delta\) with \(A_1\) to represent that the second simple harmonic motion has a phase difference of \(\delta\) with the first simple harmonic motion.

The resultant \(\vec{A}\) of these two vectors will represent the resultant simple harmonic motion. As we know from vector algebra, the magnitude of the resultant vector is
\(
A=\sqrt{A_1^2+2 A_1 A_2 \cos \delta+A_2^2}
\)
which is same as equation (12.27). The resultant \(\vec{A}\) makes an angle \(\varepsilon\) with \(\overrightarrow{A_1}\), where
\(
\tan \varepsilon=\frac{A_2 \sin \delta}{A_1+A_2 \cos \delta}
\)
which is same as equation (12.28).
This method can easily be extended to more than two vectors. Figure (12.17) shows the construction for adding three simple harmonic motions in the same direction.

\(
\begin{aligned}
& x_1=A_1 \sin \omega t \\
& x_2=A_2 \sin \left(\omega t+\delta_1\right) \\
& x_3=A_3 \sin \left(\omega t+\delta_2\right) .
\end{aligned}
\)
The resultant motion is given by \(x=A \sin (\omega t+\varepsilon)\).

Composition of Two Simple Harmonic Motions in Perpendicular Directions

Suppose two forces act on a particle, the first alone would produce a simple harmonic motion in \(x\)-direction given by
\(
x=A_1 \sin \omega t \dots(i)
\)
and the second would produce a simple harmonic motion in \(y\)-direction given by
\(
y=A_2 \sin (\omega t+\delta) \dots(ii)
\)
The amplitudes \(A_1\) and \(A_2\) may be different and their phases differ by \(\delta\). The frequencies of the two simple harmonic motions are assumed to be equal. The resultant motion of the particle is a combination of the two simple harmonic motions. The position of the particle at time \(t\) is \((x, y)\) where \(x\) is given by equation (i) and \(y\) is given by (ii). The motion is thus two-dimensional and the path of the particle is in general an ellipse. The equation of the path may be obtained by eliminating \(t\) from (i) and (ii).
By (i),
Thus, \(\quad \cos \omega t=\sqrt{1-\frac{x^2}{A_1{ }^2}}\).
Putting in (ii)
\(
y=A_2[\sin \omega t \cos \delta+\cos \omega t \sin \delta]
\)
\(=A_2\left[\frac{x}{A_1} \cos \delta+\sqrt{1-\frac{x^2}{A_1^2}} \sin \delta\right]\)
or
\(\left(\frac{y}{A_2}-\frac{x}{A_1} \cos \delta\right)^2=\left(1-\frac{x^2}{A_1^2}\right) \sin ^2 \delta\)
or, \(\quad \frac{y^2}{A_2^2}-\frac{2 x y}{A_1 A_2} \cos \delta+\frac{x^2}{A_1^2} \cos ^2 \delta\)
\(=\sin ^2 \delta-\frac{x^2}{A_1^2} \sin ^2 \delta\)
or, \(\frac{x^2}{A_1^2}+\frac{y^2}{A_2^2}-\frac{2 x y \cos \delta}{A_1 A_2}=\sin ^2 \delta \dots(12.29)\).
This is an equation of an ellipse and hence the particle moves in ellipse. Equation (i) shows that \(x\) remains between \(-A_1\) and \(+A_1\) and (ii) shows that \(y\) remains between \(A_2\) and \(-A_2\). Thus, the particle always remains inside the rectangle defined by
\(
x= \pm A_1, y= \pm A_2 \text {. }
\)
The ellipse given by (12.29) is traced inside this rectangle and touches it on all the four sides (figure 12.18).

Special Cases (\(\delta=0\))

The two simple harmonic motions are in phase. When the \(x\)-coordinate of the particle crosses the value 0, the \(y\)-coordinate also crosses the value 0. When \(x\)-coordinate reaches its maximum value \(A_1\), the \(y\)-coordinate also reaches its maximum value \(A_2\). Similarly, when \(x\)-coordinate reaches its minimum value \(-A_1\), the \(y\)-coordinate reaches its minimum value \(-A_2\).

If we substitute \(\delta=0\) in equation (12.29) we get
\(
\frac{x^2}{A_1^2}+\frac{y^2}{A_2^2}-\frac{2 x y}{A_1 A_2}=0
\)
or,
\(
\left(\frac{x}{A_1}-\frac{y}{A_2}\right)^2=0
\)
or,
\(
y=\frac{A_2}{A_1} x \dots(iii)
\)
which is the equation of a straight line passing through the origin and having a slope \(\tan ^{-1} \frac{A_2}{A_1}\). Figure (12.19) shows the path. Equation (iii) represents the diagonal \(A C\) of the rectangle. The particle moves on this diagonal.

Equation (iii) can be directly obtained by dividing (i) by (ii) and putting \(\delta=0\). The displacement of the particle on this straight line at time \(t\) is
\(
\begin{aligned}
r=\sqrt{x^2+y^2} & =\sqrt{\left(A_1 \sin \omega t\right)^2+\left(A_2 \sin \omega t\right)^2} \\
& =\sqrt{\left(A_1^2+A_2^2\right)} \sin \omega t .
\end{aligned}
\)
Thus, the resultant motion is a simple harmonic motion with same frequency and phase as the component motions. The amplitude of the resultant simple harmonic motion is \(\sqrt{A_1^2+A_2^2}\) as is also clear from the figure (12.19).

Special Cases (\(\delta=\pi\))

The two simple harmonic motions are out of phase in this case. When the \(x\)-coordinate of the particle reaches its maximum value \(A_1\), the \(y\)-coordinate reaches its minimum value \(-A_2\). Similarly, when the \(x\)-coordinate reaches its minimum value \(-A_1\), the \(y\)-coordinate takes its maximum value \(A_2\).
Putting \(\delta=\pi\) in equation (12.29) we get
\(
\frac{x^2}{A_1^2}+\frac{y^2}{A_2^2}+\frac{2 x y}{A_1 A_2}=0
\)
or,
\(
\left(\frac{x}{A_1}+\frac{y}{A_2}\right)^2=0
\)
or,
\(
y=-\frac{A_2}{A_1} \cdot x
\)
\(
\text { which is the equation of the line } B D \text { in figure (12.20). }
\)

Thus the particle oscillates on the diagonal \(B D\) of the rectangle as shown in figure (12.20).
The displacement on this line at time \(t\) may be obtained from equations (i) and (ii) (with \(\delta=\pi\) ).
\(
\begin{aligned}
r & =\sqrt{x^2+y^2}=\sqrt{\left[A_1 \sin \omega t\right]^2+\left[A_2 \sin (\omega t+\pi)\right]^2} \\
& =\sqrt{A_1^2 \sin ^2 \omega t+A_2^2 \sin ^2 \omega t}=\sqrt{A_1^2+A_2^2} \sin \omega t .
\end{aligned}
\)
Thus the resultant motion is a simple harmonic motion with amplitude \(\sqrt{A_1^2+A_2^2}\).

Special Cases (\(\delta=\pi/2\))

The two simple harmonic motions differ in phase by \(\pi / 2\). Equations (i) and (ii) may be written as
\(
\begin{aligned}
& x=A_1 \sin \omega t \\
& y=A_2 \sin (\omega t+\pi / 2)=A_2 \cos \omega t .
\end{aligned}
\)

The \(x\)-coordinate takes its maximum value \(x=A_1\) when \(\sin \omega t=1\). Then \(\cos \omega t=0\) and hence, the \(y\)-coordinate is zero. The particle is at the point \(E\) in figure (12.21). When \(x\)-coordinate reduces to 0 , \(\sin \omega t=0\), and \(\cos \omega t\) becomes 1 . Then \(y\)-coordinate takes its maximum value \(A_2\) so that the particle reaches the point \(F\). Then \(x\) reduces to \(-A_1\) and \(y\) becomes 0 . This corresponds to the point \(G\) of the figure (12.21). As \(x\) increases to 0 again, \(y\) takes its minimum value \(-A_2\), the particle is at the point \(H\). The motion of the particle is along an ellipse EFGHE inscribed in the rectangle shown. The major and the minor axes of the ellipse are along the \(X\) and \(Y\)-axes.
Putting \(\delta=\pi / 2\) in equation (12.29) we get
\(
\frac{x^2}{A_1^2}+\frac{y^2}{A_2^2}=1
\)
which is the standard equation of an ellipse with its axes along \(X\) and \(Y\)-axes and with its centre at the origin. The length of the major and minor axes are \(2 A_1\) and \(2 A_2\).

If \(A_1=A_2=A\) together with \(\delta=\pi / 2\), the rectangle of figure (12.21) becomes a square and the ellipse becomes a circle. Equation (12.29) becomes
\(
x^2+y^2=A^2
\)
which represents a circle.
Thus, the combination of two simple harmonic motions of equal amplitude in perpendicular directions differing in phase by \(\pi / 2\) is a circular motion.
The circular motion may be clockwise or anticlockwise, depending on which component leads the other.

Example 2: Displacement versus time curve for a particle executing S.H.M. is shown in Figure below. Identify the points marked at which (i) the velocity of the oscillator is zero, (ii) the speed of the oscillator is maximum.

Solution:

In the displacement-time graph of SHM, zero displacement values correspond to the mean position; where the velocity of the oscillator is maximum. Whereas the crest and troughs represent amplitude positions, where displacement is maximum and velocity of the oscillator is zero.
(i) The points A, C, E, and G lie at extreme positions (maximum displacement, \(y=\) A). Hence the velocity of the oscillator is zero.
(ii) The points B, D, F, and H lie at the mean position (zero displacement, \(y=0\) ). We know the speed is maximum at the mean position.

Example 3: What are the two basic characteristics of a simple harmonic motion?

Solution:

The two basic characteristics of a simple harmonic motion
(i) Acceleration is directly proportional to displacement.
(ii) The direction of acceleration is always towards the mean position, which is opposite to displacement.

Example 4: When will the motion of a simple pendulum be simple harmonic?

Solution:

Simple pendulum performs angular S.H.M. Consider the bob of a simple pendulum is displaced through an angle \(\theta\) shown. Q
The restoring torque about the fixed point \(\mathrm{O}\) is
\(\tau=m g l \sin \theta\)
If \(\theta\) is small angle in radians, then \(\sin \theta=0\)
\(\Rightarrow \mathrm{mgl} \theta\)
In vector form \(\tau \propto \theta\)
Hence, the motion of a simple pendulum is SHM for small angle of oscillations.

Example 5: In the figure below, what will be the sign of the velocity of the point \(\mathrm{P}^{\prime}\), which is the projection of the velocity of the reference particle \(\mathrm{P}\). \(\mathrm{P}\) is moving in a circle of radius \(R\) in anticlockwise direction.

Solution:

At time \(=t, P^{\prime}\) is the foot perpendicular of the velocity vector of the particle \(P\).
This foot shifts from \(P^{\prime}\) to \(Q\), i.e., towards the negative axis, when the particle moves from \(P\) to \(P_1\).
Thus, there is -ve sign of \(\phi\) motion of \(\mathrm{P}^{\prime}\).

Note: As the particle on the reference circle moves in the anti-clockwise direction. The projection will move from \(\mathrm{P}^{\prime}\) to O towards the left, i.e. from right to left, hence sign is negative.

Example 6: Show that for a particle executing S.H.M, velocity and displacement have a phase difference of \(\pi / 2\).

Solution:

Let the displacement equation of SHM
\(
\begin{aligned}
& \qquad x=a \cos \omega t \\
& \text { velocity } v=\frac{d x}{d t}=a \omega(-\sin \omega t)=-a \omega \sin \omega t \\
& \Rightarrow \quad v=a \omega \cos \left(\frac{\pi}{2}+\omega t\right)
\end{aligned}
\)
Now, phase of displacement \(\phi_1=\omega t\)
Phase of velocity \(\phi_2=\frac{\pi}{2}+\omega t\)
\(\therefore\) Difference in the phase of velocity to that of the phase of displacement
\(
\Delta \phi=\phi_2-\phi_1=\left(\frac{\pi}{2}+\omega t\right)-(\omega t)=\frac{\pi}{2}
\)

Example 7: Draw a graph to show the variation of P.E., K.E., and total energy of a simple harmonic oscillator with displacement.

Solution:

The potential energy (PE) of a simple harmonic oscillator is
\(
\mathrm{PE}=\frac{1}{2} k x^2=\frac{1}{2} m \omega^2 x^2 \dots(i)
\)
When, \(\mathrm{PE}\) is plotted against displacement \(x\), we will obtain a parabola.
When \(x=0, \mathrm{PE}=0\).
When \(x= \pm A, \mathrm{PE}=\) maximum
\(
=\frac{1}{2} m \omega^2 A^2
\)
The kinetic energy \((\mathrm{KE})\) of a simple harmonic oscillator \(\mathrm{KE}=\frac{1}{2} m v^2\) But velocity of oscillator \(v=\omega \sqrt{A^2-x^2}\)
\(
\Rightarrow \quad \mathrm{KE}=\frac{1}{2} m\left[\omega \sqrt{A^2-x^2}\right]^2
\)
or \(\quad \mathrm{KE}=\frac{1}{2} m \omega^2\left(A^2-x^2\right) \dots(ii)\)
This is also a parabola, if we plot KE against displacement \(x\).
i.e., \(\quad \mathrm{KE}=0\) at \(x= \pm A\)
and \(\quad \mathrm{KE}=\frac{1}{2} m \omega^2 A^2\) at \(x=0\)
Now, the total energy of the simple harmonic oscillator \(=\mathrm{PE}+\mathrm{KE}\) [using Eqs. (i) and (ii)]
\(
\begin{aligned}
& =\frac{1}{2} m \omega^2 x^2+\frac{1}{2} m \omega^2\left(A^2-x^2\right) \\
& =\frac{1}{2} m \omega^2 x^2+\frac{1}{2} m \omega^2 A^2-\frac{1}{2} m \omega^2 x^2 \\
\mathrm{TE} & =\frac{1}{2} m \omega^2 A^2=\text { constant }
\end{aligned}
\)
which is a constant and independent of \(x\).
Plotting under the above guidelines \(\mathrm{KE}, \mathrm{PE}\) and \(T E\) versus displacement \(x\)-graph as follows:

Important point: From the graph, we note that potential energy or kinetic energy completes two vibrations in a time during which S.H.M. completes one vibration. Thus the frequency of potential energy or kinetic energy is double that of S.H.M.

Example 8: Show that the motion of a particle represented by \(y=\sin \omega t-\cos \omega t\) is simple harmonic with a period of \(2 \pi / \omega\).

Solution:

The given equation is in the form of a combination of two harmonic functions. We can write this equation in the form of a single harmonic (sine or cosine) function.
We have displacement function: \(y=\sin \omega t-\cos \omega t\)
We have displacement function: \(y=\sin \omega t-\cos \omega t\)
\(
\begin{aligned}
y & =\sqrt{2}\left(\frac{1}{\sqrt{2}} \cdot \sin \omega t-\frac{1}{\sqrt{2}} \cdot \cos \omega t\right) \\
& =\sqrt{2}\left[\cos \left(\frac{\pi}{4}\right) \cdot \sin \omega t-\sin \left(\frac{\pi}{4}\right) \cdot \cos \omega t\right] \\
& =\sqrt{2}\left[\sin \left(\omega t-\frac{\pi}{4}\right)=\sqrt{2}\left[\sin \left(\omega t-\frac{\pi}{4}\right)\right]\right] \\
\Rightarrow \quad y & =\sqrt{2} \sin \left(\omega t-\frac{\pi}{4}\right)
\end{aligned}
\)
Comparing with the standard equation of S.H.M,
\(
y=a \sin (\omega t+\phi) \text { we get angular frequency of S.H.M, } \omega=\frac{2 \pi}{T} \Rightarrow T=\frac{2 \pi}{\omega}
\)
Hence the function represents SHM with a period \(T=2 \pi / \omega\)

Example 9: A person normally weighing \(50 \mathrm{~kg}\) stands on a massless platform that oscillates up and down harmonically at a frequency of \(2.0 \mathrm{~s}^{-1}\) and an amplitude \(5.0 \mathrm{~cm}\). A weighing machine on the platform gives the person’s weight against time.
(a) Will there be any change in the weight of the body, during the oscillation?
(b) If the answer to part (a) is yes, what will be the maximum and minimum reading in the machine, and at which position?

Solution:

(a) Yes.
(b) Maximum weight \(=M g+M A \omega^2\)
\(
\begin{aligned}
& =50 \times 9.8+50 \times \frac{5}{100} \times(2 \pi \times 2)^2 \\
& =490+400=890 \mathrm{~N} .
\end{aligned}
\)
\(
\begin{aligned}
& \text { Minimum weight }=M g-M A \omega^2 \\
& \\
& =50 \times 9.8-50 \times \frac{5}{100} \times(2 \pi \times 2)^2 \\
& =490-400 \\
& =90 \mathrm{~N} .
\end{aligned}
\)
Maximum weight is at the topmost position,
Minimum weight is at the lowermost position.

Example 10: A cylindrical log of wood of height \(h\) and area of cross-section \(A\) floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.
\(
T=2 \pi \sqrt{\frac{m}{A \rho g}}
\)
where \(m\) is mass of the body and \(\rho\) is density of the liquid.

Solution:

Let the \(\log\) be pressed and let the vertical displacement at the equilibrium position be \(x_0\).
At equilibrium
\(
\begin{aligned}
m g & =\text { Buoyant force } \\
& =A x_o \rho g
\end{aligned}
\)
When it is displaced by a further displacement \(x\), the buoyant force is \(A\left(x_o+x\right) \rho g\).
Net restoning force
\(
\begin{aligned}
& =\text { Buoyant force }- \text { weight } \\
& =A\left(x_o+x\right) \rho g-m g \\
& =(A \rho g) x \text {. i.e. proportional to } x \\
\therefore T=2 \pi & \sqrt{\frac{m}{A \rho g}}
\end{aligned}
\)

Example 11: One end of a V-tube containing mercury is connected to a suction pump and the other end to the atmosphere. The two arms of the tube are inclined to horizontally at an angle of \(45^{\circ}\) each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in the V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

Solution:

Let \(l\) be the length of liquid in each arm of \(\mathrm{V}\) tube in equilibrium position. Let \(\mathrm{A}\) be the area of cross-section of the liquid and \(\rho\) be the density of the liquid.
Mass of liquid in V-tube, \(m=2l \mathrm{~A} \rho\)
Let the liquid be pushed down through a small distance \(\mathrm{x}\) is one arm of \(\mathrm{V}\) tube.
Its level will be raised by distance \(\mathrm{x}\) in other arm of \(\mathrm{V}\) tube.
Pressure difference due to difference of levels in \(\mathrm{V}\) tube is
\(
\Delta \mathrm{P}=2 {x}\left(\sin 45^{\circ}\right) \mathrm{\rho g}
\)
Thrust perpendicular to the liquid meniscus in the tube is
\(
F=-\frac{(\Delta P) A}{\cos 45^{\circ}}=-2 \rho g A x \dots(i)
\)
Now \(F \propto(-x)\)
So, it will show SHM
So, \(\mathrm{k}=2 \rho \mathrm{gA}\)
Time period \(T=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=2 \pi \sqrt{\frac{2 l A \rho}{2 \rho g A}}=2 \pi \sqrt{\frac{l}{g}}\)

Example 12: A tunnel is dug through the centre of the Earth. Show that a body of mass ‘ \(m\) ‘ when dropped from rest from one end of the tunnel will execute the simple harmonic motion.

Solution:

Acceleration due to gravity at \(P=\frac{g \cdot x}{R}\), where \(g\) is the acceleration at the surface.
Force \(=\frac{m g x}{R}=-k \cdot x, \quad k=\frac{m g}{R}\)
Motion will be SHM with time period \(T=\sqrt{\frac{m}{K}}=2 \pi \sqrt{\frac{R}{g}}\)

Example 13: A simple pendulum of time period 1s and length \(l\) is hung from a fixed support at \(\mathrm{O}\), such that the bob is at a distance H vertically above A on the ground (figure below). The amplitude is \(\theta_0\). The string snaps at \(\theta=\theta_0 / 2\). Find the time taken by the bob to hit the ground. Also, find the distance from A where bob hits the ground. Assume \(\theta_0\) to be small so that \(\sin \theta_0 = \theta_0\) and \(\cos \theta_0 =1\).

Solution:

Let us assume \(t=0\) when \(\theta_0=\theta_0\), then \(\theta=\theta_0 \cos \omega t\)
Given a seconds pendulum \(\omega=2 \pi\)
\(
\Rightarrow \theta=\theta_0 \cos 2 \pi \mathrm{t} \dots(i)
\)
At time \(t_1\), let \(\theta=\theta_0 / 2\)
\(
\begin{aligned}
& \therefore \cos 2 \pi \mathrm{t}_1=1 / 2 \\
& \Rightarrow t_1=\frac{1}{6} \ldots \ldots .\left[\because \cos 2 \pi t_1=\cos \frac{\pi}{3}=2 \pi t_1=\frac{\pi}{3}\right] \\
& \frac{d \theta}{d t}=-\left(\theta_0 2 \pi\right) \sin 2 \pi t \ldots . .[\text { From equation (i)] } \\
& \text { At } t=t_1=\frac{1}{6} \\
& \frac{d \theta}{d t}=-\theta_0 2 \pi \sin \frac{2 \pi}{6}=-\sqrt{3} \pi \theta_0
\end{aligned}
\)
A negative sign shows that it is going left.
Thus, the linear velocity is \(u=-\sqrt{3} \pi \theta_0 l\) perpendicular to the string.
The vertical component is \(u_y=-\sqrt{3} \pi \theta_0 l \sin \left({\theta_0}\right)\)
And the horizontal component is \(u_x=-\sqrt{3} \pi \theta_0 l \cos \left({\theta_0}\right)\)
At the time it snaps, the vertical height is \(H^{\prime}=H+l\left(1-\cos \left(\theta_0 / 2\right)\right)\)
Let the time required for fall be \(t\), then
\(H^{\prime}=u_y t+(1 / 2) g t^2\) (notice \(g\) is also in the negative direction)

Or, \(\frac{1}{2} g t^2+\sqrt{3} \pi \theta_0 l \sin \theta_0 t-H^{\prime}=0\) 

\(
\begin{aligned}
& \therefore t=\frac{-\sqrt{3} \pi \theta_0 l \sin \theta_0 \pm \sqrt{3 \pi^2 \theta_0^2 \mathrm{e}^2 \sin ^2 \theta_0+2 g H^{\prime}}}{g} \\
& \quad \frac{-\sqrt{3} \pi l \theta_0^2 \pm \sqrt{3 \pi^2 \theta_0^4 l^2+2 g H^{\prime}}}{g}
\end{aligned}
\)
Neglecting terms of order \(\theta_0^2\) and higher,
\(t = \sqrt{\frac{2 H^{\prime}}{g}}\).
\(
\text { Now } H^{\prime}=\quad H+l(1-1)=H
\)
\(
\therefore t =\sqrt{\frac{2 H}{g}}
\)
The distance travelled in the \(x\) direction is \(u_x t\) to the left of where it snapped.
\(
X=\sqrt{3} \pi \theta_0 l \cos \theta_0 \sqrt{\frac{2 H}{g}}
\)
To order of \(\theta_0\)
\(
X=\sqrt{3} \pi \theta_0 l \sqrt{\frac{2 H}{g}}=\sqrt{\frac{6 H}{g}} \theta_0 l .
\)
At the time of snapping, the bob was
\(l \sin \theta_0 =\quad l \theta_0\) distance from \(\mathrm{A}\)
Thus, the distance from \(A\) is
\(
l \theta_o-\sqrt{\frac{6 H}{g}} l \theta_o=l \theta_o(1-\sqrt{6 H / g})
\)